We want to find the of the given .
$y=(x+1)(x−1)(x+2) $
Zeros of the Function
To do so, we need to find the values of
$x$ for which
$y=0.$
$y=0⇔(x+1)(x−1)(x+2)=0 $
Since the function is already written in , we will use the .
$(x+1)(x−1)(x+2)=0$
$x+1=0x−1=0x+2=0 (I)(II)(III) $
$x=−1x−1=0x+2=0 (I)(II)(III) $
$x=−1x=1x+2=0 (I)(II)(III) $
$x=−1x=1x=−2 (I)(II)(III) $
We found that the zeros of the function are
$2,$ $1,$ and
$1.$ Graph
To draw the graph of the function, we will find some additional points and consider the . Let's use a table to find additional points.
$x$

$(x+1)(x−1)(x+2)$

$y=(x+1)(x−1)(x+2)$

$3$

$(3+1)(3−1)(3+2)$

$8$

$1.5$

$(1.5+1)(1.5−1)(1.5+2)$

$0.625$

$0$

$(0+1)(0−1)(0+2)$

$2$

$0.5$

$(0.5+1)(0.5−1)(0.5+2)$

$1.875$

The points
$(3,8),$ $(1.5,0.625),$ $(0,2),$ and
$(0.5,1.875)$ are on the graph of the function. Finally, let's apply the . This will simplify the equation and determine the leading and of the polynomial function.
$y=(x+1)(x−1)(x+2)$
$y=(x(x−1)+1(x−1))(x+2)$
$y=(x_{2}−x+1(x−1))(x+2)$
$y=(x_{2}−x+x−1)(x+2)$
$y=(x_{2}−1)(x+2)$
$y=x_{2}(x+2)−1(x+2)$
$y=x_{3}+2x_{2}−1(x+2)$
$y=x_{3}+2x_{2}−x−2$
We can see now that the leading coefficient is
$1,$ which is a positive number. Also, the degree is
$3,$ which is an odd number. Therefore, the end behavior is
down and
up.
The function correpsonds to graph $B.$